Mathcad P-elements linear versus nonlinear stress 2014-t6

P-Elements Linear Versus Nonlinear Stress 2014-T6.xmcd

Nonlinear analytical solution based on linear maximum stress at a hole for 2014-T6Aluminum Alloy

ORIGIN 1≡

References1. MMPDS-03, 1 October of 2006 2. ESDU 76016

Geometry of the hole: P 0.4375− 0.1660− 0.0880− 0.0100− 0.4375( )T in⋅≡

Far field traction

σg 36.428 ksi⋅≡

Maximum Edge Distance: e P5

P3

− 0.525 in=≡

Minimum Edge Distance: c P3

P1

− 0.349 in=≡

Plate Width: W e c+ 0.875 in=:=

Hole Diameter: D P4

P2

− 0.156 in=≡

Hole Radius: rD

20.078 in=≡

Nominal-Stress Path, from a to b: a r 0.078 in=:=

b c 0.349 in=:=

Dimensionless Hole location: ψe

c1.504=:=

Dimensionless Edge short-edge Distance: λD

2 c⋅0.2232=:=

Hole-Location Eccentricity: δψ 1−

2 ψ 1+( )⋅

W⋅ 0.088 in=:=

Julio C. Banks, P.E. page 1 of 23


Stress-strain Curve for 2014-T6 Aluminum Extrusions for t 0.500 in⋅≥

Fundamental Properties at room temperature per Reference 1, page 3-50

Modulus of Elasticity: E 10.8 103⋅ ksi⋅≡

Ultimate Tensile Stress: Ftu 64 ksi⋅≡

Yield Tensile Stress: Fty 58 ksi⋅≡

Ultimate Elongation: eu 7 %⋅≡

Poisson's Ratio: μ 0.33:= Only required for FEM solution

n = f(T): n 26≡ Reference 1, page 3-69

The following parameters are calculated in Appendix B

MMPDS Proportional Limit: SPL 53.08 ksi⋅= εPL 0.512 %⋅=

SPL

S0.291.5 %⋅=

StressCheck Required: S70E 58.327 ksi⋅= ε70E 0.772 %⋅=

Notice that the ultimate strain, eu, is measured after the test specimen has failed and there

is no load. Therefore, the actual ultimate strain at the ultimate load, Ftu 64 ksi⋅= , is:

εu 7.59 %⋅=



Figure 1: MMPDS-03 Stress Strain for Aluminum Alloy Extrusion 2014-T6



Perform a FEM study using the P-element formulation used in Stress Check forψ 1.504= and the Hole-Location Eccentricity δ 0.088 in=

Figure 2: Material parameters StressCheck requires to generate the Ramberg-OsgoodStress-strain data set



Figure 3: Linear stress solution using P-element StressCheck FEA Software



Figure 4: Nonlinear stress solution using P-element StressCheck FEA Software



FEM from StressCheck

Linear Solution Nonlinear Solution

Maximum stress: σMax max σFEM Y( )( ) 116 ksi⋅=:= σ'Max max σ'FEM Y( )( ) 61.66 ksi⋅=:=

Short-edge stress: σNoma

b

yσFEM y( )⌠⌡

d

b a−44.59 ksi⋅=:= σ'Nom

a

b

yσ'FEM y( )⌠⌡

d

b a−44.05 ksi⋅=:=

Ultimate Stress: σu 1.50σNom⋅ 66.88 ksi⋅=:= σ'u 1.50σ'Nom⋅ 66.08 ksi⋅=:=

Gross SCF: Ktg_FEM

σMax

σg3.184=:= K'tg_FEM

σ'Max

σg1.693=:=

Nominal SCF: Ktn_FEM

σMax

σNom2.602=:= K'tn_FEM

σ'Max

σ'Nom1.400=:=

Nominal to gross transformation factor:

ΛFEM

Ktg_FEM

Ktn_FEM1.2240=:= Λ'FEM

K'tg_FEM

K'tn_FEM1.2093=:=

ΛFEM

σNom

σg1.2240=:= Λ'FEM

σ'Nom

σg1.2093=:=

MSu

Ftu

σu1− 0.04−=:= MS'u

Ftu

σ'u1− 0.03−=:=

Closed-form Solution

λ 0.2232= K'tg λ ψ, ( ) 3.184= σMax_CF K'tg λ ψ, ( ) σg⋅ 116.0 ksi⋅=:=

ψ 1.504= K'tn λ ψ, ( ) 2.481= ΛCF

K'tg λ ψ, ( )

K'tn λ ψ, ( )1.2830=:= or Λ λ ψ, ( ) 1.276=

The maximum strain found in Figure 5 is to be multiplied by 1.5 to obtain the ultimate strain toallow the calculation of the Margin of Safety based on strain rather than stress as follows

StressCheck FEM Maximum Nonlinear Strain: εMax_SC 1.757 %⋅:=

StressCheck Margin of Safety based on Strain: MSε

eu

1.5εMax_SC⋅1− 1.66=:=



0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.430

40

50

60

70

80

90

100

110

120

Linear VonMises Stress, ksiNolinear VonMises Stress, ksi

Von Mises Stress Vs Lateral Location in a Plate with Eccentric Hole

σ'Nom

ksi

σ'u

ksi

σFEM Y( )

ksi

σ'FEM Y( )

ksi

r

in

Y

in



Determine a nonlinear Margin of Safety based upon the maximum strain at the hole dueto the negative Margin of Safety based on stress.

Figure 5: Nonlinear strain solution using P-element StressCheck FEA Software



0.0 0.3 0.5 0.8 1.0 1.3 1.5 1.8 2.0 2.3 2.5 2.8 3.0 3.3 3.50.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

Ramberg-Osgood EquationRamberg-Osgood Equation Generated by StressCheck70% E Curve

Ramberg-Osgood Equation for Aluminum 2014-T6

Strain, %

Str

ess,

ksi

SPL

ksi

Ftu

ksi

εPL

%

ε70E

%

MMPDS 03 StressCheck Relative Error

SPL 53.084 ksi⋅= S'PL σSC εPL( ) 50.755 ksi⋅=:= %∆ SPL S'PL, ( ) 4.4− %⋅=

S70E 58.327 ksi⋅= S'70E σSC ε70E( ) 57.918 ksi⋅=:= %∆ S70E S'70E, ( ) 0.7− %⋅=

Su_RO 70.774 ksi⋅= S'u_RO σSC εu( ) 66.491 ksi⋅=:= %∆ Su_RO S'u_RO, ( ) 6.1− %⋅=

Note εu 7.593 %⋅= is the strain obtained from Ramberg-Osgood equation Corresponding to the loaded

test specimen. The actual ultimate strain provided in MMPDS (and similar reference books) is which is a measured strain of the ruptured (unloaded) test specimen



Apply the ESED (Equivalent Strain-Energy Density also known as Glinka) Method to obtain thepseudo nonlinear solution using the linear FEM maximum stress at the short-edge of a hole in a plate in a tensile stress-field.

Strain-energy constant: Γ1

2εL⋅ σFEM⋅= 1a( )

σFEM E0 εL⋅= 2( )

Solve for the linear strain, εL from (2) as

εL

σFEM

E0= 3( )

Substitute (3) into (1) to get Γ1

2

σFEM2

E0⋅= 1b( )

Determine the strain at which the Ramberg-Osgood equates the strain-energy constant, Γ (Equation 1b)

The linear FEM Maximum Stress Using P-elements (StressCheck) is σFEM 116 ksi⋅≡

The strain-energy for the Given σFEM is Γ1

2

σFEM2

E⋅ 622.96 psi=:=

Initial guess: ε ε0.2 0.737 %⋅=:=

Given

Γ

0

ε

εσRO_rootε( )⌠⌡

d=

εG Find ε( ):=

The Pseudo Non-linear Strain is εG 1.289 %⋅=



Post-process

σGlinka σRO_rootεG( ) 65.466 ksi⋅=:=The Pseudo Non-linear Strain is

The P-element (StressCheck) Maximum Non-linear Strain is εMax_SC 1.757 %⋅=

The ESED (Glinka) Maximum Pseudo Non-linear Strain is εG 1.289 %⋅=

Relative Error of the ESED strain solution with respect to the P-element (FEM) solution is

%∆ εMax_SC εG, ( ) 26.66− %⋅=

StressCheck Margin of Safety based on Strain: MSε

eu

1.5εMax_SC⋅1− 1.66=:=

ESED Margin of Safety based on Strain: MSu

eu

1.50εG⋅1− 2.62=:=

The P-element (StressCheck) Maximum Non-linear Stress is σ'Max 61.66 ksi⋅=

The ESED (Glinka) Maximum Pseudo Non-linear Stress is σGlinka 65.466 ksi⋅=

Relative Error of the ESED stress solution with respect to the P-element (FEM) solution is

%∆ σ'Max σGlinka, ( ) 6.17 %⋅=

ConclusionThe ESED (Equivalent-strain Energy Density) also known as Glinka can be used to obtain anestimate of the plastic strain and corresponding MS (Margin of Safety) when the linear solutionproduces a negative MS indicating nonlinear material behavior due to high loads.

It should be noted that this approach will most likely occur for the ultimate load Marginof Safety, MSu, rather than the limit load (yield) margin of safety, MSY



Appendix A: StressCheck P-element FEM Solution Data and Cubic Spline Procedure

Ramberg-Osgood Numerical Results from StressCheck

Linear Solution Data Nonlinear Solution Data

DataSC

0.0000000

0.0043478

0.0086957

0.0130430

0.0173910

0.0217390

0.0260870

0.0304350

0.0347830

0.0391300

0.0434780

0.0478260

0.0521740

0.0565220

0.0608700

0.0652170

0.0695650

0.0739130

0.0782610

0.0826090

0.0869570

0.0913040

0.0956520

0.1000000

0

46872

59076

60992

62066

62818

63397

63870

64269

64615

64920

65193

65440

65666

65875

66068

66248

66416

66575

66724

66866

67001

67129

67251

≡

Data

0.1660−

0.1725−

0.1789−

0.1854−

0.1919−

0.1983−

0.2048−

0.2113−

0.2177−

0.2242−

0.2306−

0.2371−

0.2436−

0.2500−

0.2565−

0.2630−

0.2694−

0.2759−

0.2824−

0.2888−

0.2953−

0.3018−

0.3244−

0.3470−

0.3696−

0.3923−

0.4149−

0.4375−

116000.0

93230.0

78340.0

68310.0

61340.0

56320.0

52600.0

49750.0

47540.0

45780.0

44380.0

43270.0

42390.0

41700.0

41170.0

40760.0

40430.0

40160.0

39920.0

39690.0

39470.0

39265.0

38910.0

38450.0

37990.0

37370.0

36260.0

34270.0

≡ Data'

0.1660−

0.1725−

0.1789−

0.1854−

0.1919−

0.1983−

0.2048−

0.2113−

0.2177−

0.2242−

0.2306−

0.2371−

0.2436−

0.2500−

0.2565−

0.2630−

0.2694−

0.2759−

0.2824−

0.2888−

0.2953−

0.3018−

0.3141−

0.3264−

0.3388−

0.3511−

0.3635−

0.3758−

0.3881−

0.4005−

0.4128−

0.4252−

0.4375−

61660

61150

60580

59930

59180

58300

57250

55940

54230

51950

49330

47070

45430

44310

43550

42990

42510

42010

41480

40920

40370

39935

39790

39540

39270

39000

38740

38440

38040

37480

36660

35510

33920

≡



FCS x X, Y, ( ) "Cubic-spline"

S cspline X Y, ( )←

y interp S X, Y, x, ( )←

≡

StressCheck Calculated Ramberg-Osgood Material Stress-strain Data

εSC_RO_Data DataSC1⟨ ⟩≡

σSC_RO_Data DataSC2⟨ ⟩ psi⋅≡

σSC ε( ) FCS ε εSC_RO_Data, σSC_RO_Data, ( )≡

StressCheck Linear solution P-elements StressCheckNonlinear solution P-elements

Y P3

Data1⟨ ⟩ in⋅−

≡ Y' P

3Data'1

⟨ ⟩in⋅−

≡

SSC Data2⟨ ⟩ psi⋅≡ S'SC Data'2⟨ ⟩

psi⋅≡

σFEM y( ) FCS y Y, SSC, ( )≡ σ'FEM y( ) FCS y Y', S'SC, ( )≡



Appendix B: Functions

Ramberg-Osgood Stress-strain equation:

The defined 0.2% offset strain is: e0.2 0.2 %⋅≡

Yield Stress: S0.2 Fty 58 ksi⋅=≡

Yield Strain: ε0.2 e0.2

S0.2

E+ 0.74 %⋅=≡

0.2% offset Linear-stress: σL ε( ) E ε e0.2−( )⋅≡

Normalization stress: σn S0.2 58 ksi⋅=≡

Equation [9.8.4.1.2(b)]Ref. 1, page 9-200εRO σ( )

σ

Ee0.2

σ

σn

n

⋅+≡

Reference 1, Section 9.8.4.6.1 (page 9-200) states:If the proportional limit stress is equated with a plastic strain level of 0.0002 or a 0.02 percentdeviation from linearity, and the Ramberg- Osgood relationship is found to be valid for smallplastic strains, then the proportional limit stress, SPL fpl= , can be approximated from Equation

9.8.4.6(a) as follows:

Section [9.8.4.6.1]Ref. 1, page 9-200

Proportional Limit: SPL S0.2 0.1( )

1

n⋅ 53.08 ksi⋅=≡

εPL εRO SPL( ) 0.512 %⋅=≡

MMPDS 70%E Data point

S70E

S0.2

ksi

7

3e0.2⋅

E

ksi⋅

1

n

n

n 1−( )

ksi⋅

58.327 ksi⋅=≡ε70E 150

S0.2

70 %⋅ E⋅

n

⋅

1

n 1−( )

0.772 %⋅=≡

Define the 0.70xE straight line which when intercepting the Stress-strain curve will produce the S70Estress level required by StressCheck as σ70E ε( ) 0.70 E⋅( ) ε⋅≡

Ultimate Stress: Su Ftu 64 ksi⋅=≡ εu eu

Ftu

E+ 7.59 %⋅=≡

Relative-Error: %∆ a b, ( )b a−

a

≡



The Ramberg-Osgood Equation [9.8.4.1.2(b)] is an implicit equation which needs to be solved using aroot-finding procedure since ε =f(σ), and not σ = g(ε). However, there is an approximation, σARO ε( ),

given in ESDU 76016, page 76016 as equation 5.1 which can be used as an estimate of the stress,σRO(ε), or even as a seed (starting value) for a more refined root-finding procedures.

σARO ε( ) "ESDU 76016, page 76016, equation 5.1"

k 0.79044 n⋅ 0.86977−←

β 11

n+

n k− 1−( )

11

n+

k−−←

λε E⋅σn

←

σEP ε E⋅ 1 β λk⋅+

1

nλ

n 1−⋅+

1−

n

⋅←

≡

Et σ( )E

1 e0.2 n⋅E

σn⋅

σ

σn

n

⋅+

≡ [9.8.4.2(b)]

Ref. 1, page 9-205

Et S0.2( ) 1.011 103× ksi⋅=

σRO_rootεRoot( ) "Root-finding Procedure by Julio C. Banks, P.E."

"Using Newton's rather than MathCAD's Method"

ε1

εRoot←

σ1

σARO ε1( )←

∆σTol 0.5 psi⋅←

∆σ 2 ∆σTol⋅←

i 1←

i i 1+←

εi

εRO σi 1−( )←

σi

σi 1− ε

iε

i 1−−( ) Et σi 1−( )⋅−←

∆σ σi

σi 1−−←

∆σ ∆σTol>while

σ σi

←

≡

Su_RO σRO_rootεu( ) 70.774 ksi⋅=≡



Create a procedure that generates a set of data points corresponding to the Ramberg-Osgood Equation(1) call such a set of discrete Data Points. Assume a linear stress up until the Proportional Limit.

RO_Points NPe 2←

ε1

0 %⋅←

σ1

0 ksi⋅←

ε2

εPL←

σ2

SPL←

∆σ 0.8 ksi⋅←

NPp roundFtu SPL−

∆σ

←

NPep NPe NPp+ 2−←

σi

σi 1− ∆σ+←

εi

εRO σi( )←

i 3 NPep( )..∈for

σNPep 1+ Ftu←

εNPep 1+ εRO Ftu( )←

Data1⟨ ⟩ε←

Data2⟨ ⟩ σ

ksi←

Data

≡

Ramberg-Osgood Discrete set of Data Points

ε RO_Points1⟨ ⟩

≡ σRO RO_Points2⟨ ⟩

ksi⋅≡

Number of points generated, NRO length ε( ) 15=≡

The Ultimate Ramberg-Osgood Strains is εu_RO εNRO

3.18 %⋅=≡



Appendix C: Determine the explicit equations for of the stress level corresponding to the interceptionpoint of the 0.70E line with the Ramberg-Osgood Curve

σ70E ε( ) 0.70 E⋅( ) ε⋅=C1( )

Ramberg-Osgood: εσ

Ee0.2

σ

S0.2

n

⋅+= Equation [9.8.4.1.2(b)]Ref. 1, page 9-200Renamed here as (C2)

Where n 1>

e0.2 0.20 %⋅=

S0.2 Fty=

Let the interception point be

ε ε0.70E=

S0.70E σ70E ε0.70E( )=

Introduce the interception point definition into equations C1 and C2 while simultaneouslyequating the strain from C1 with that of the Ramberg-Osgood equation C2

S0.70E

0.70 E⋅( )

S0.70E

Ee0.2

S0.70E

S0.2

n

⋅+=



S0.70E

0.70 E⋅( )

S0.70E

Ee0.2

S0.70E

S0.2

n

⋅+=

1

0.701−

S0.70E

E⋅ e0.2

S0.70E

S0.2

n

⋅=

3

7

S0.70E

E⋅ e0.2

S0.70E

S0.2

n

⋅=

Since n 1> , then divide through by S0.70E( ) e0.2( )⋅

3

7 e0.2⋅

1

E⋅

S0.70E( ) n 1−( )

S0.2n

=

Solve for S0.70E : S0.70E

S0.2

7 e0.2⋅

3

E⋅

1

n

n

n 1−

=

Recall that e0.2 0.2 %⋅= therefore, 7 e0.2⋅ 0.014= , that is

S0.70E

S0.2

0.014

3

E⋅

1

n

n

n 1−

=3( )



Determine the strain, at which the 0.70xE line, C1, intercepts the Ramberg-Osgood line, C2

ε70E

0.70 E⋅( ) ε70E⋅

Ee0.2

0.70 E⋅( ) ε70E⋅

S0.2

n

⋅+=

0.30ε70E⋅ e0.2

0.70 E⋅( ) ε70E⋅

S0.2

n

⋅=

Divide through by ε70E( ) e0.2( )⋅ 0.30

e0.2

0.70 E⋅S0.2

n

ε70En 1−( )⋅=

Solve for ε70E ε70E0.30

e0.2

S0.2

0.70 E⋅

n

⋅

1

n 1−

=

Again, e0.2 0.2 %⋅= therefore, 0.30

e0.2150= , that is

ε70E 150S0.2

0.70 E⋅

n

⋅

1

n 1−

= 4( )

The procedure for obtaining the explicit S0.70E from a cubic-spline stress-strain data set, or a

stress-strain model such as equation C2 (Ramberg-Osgood) is as follows:

Step 1: Solve ε70E for the given the yield strength, S0.2 Fty= , the modulus of elasticity, E, and

the exponent, n

Step 2: S0.70E FG ε70E( )=

Where the subscript of the function may be either RO, for Ramberg-Osgood, or CS for Cubic-spline interpolation function

It should be noted that if step 1 is not followed, then a root-finding procedure must be utilizedto numerically find the interception point, ε ε70E=



Appendix D: Stress Concentration Factor (SCF) of a plate in tension with an eccentric hole.

Figure 1: Eccentric circular hole in finite-width plane Figure 2: Axial tension

Yield-stress Margin of Safety: MSy

Fty

σnom1−=

Ultimate-stress Margin of Safety: MSy

Ftu

1.50σnom⋅1−= or MSy

Ftu

σult1−=

Where σnom Λ λ ψ, ( ) σg⋅=B

A

yσ y( )⌠⌡

d

A B−= λ

d

2 c⋅= and ψ

e

c=

Peterson's Function (Reference 1): α 3.000 3.140− 3.667 1.527−( )T:=

Kn λ( ) α1

2

4

i

αiλ

i 1−⋅

∑

=

+:=1( )

2( )Λ λ ψ, ( )

1 λ2−

1 λ−( )

1

11

ψ1 1 λ

2−−( )⋅−

⋅≡

Kt λ ψ, ( ) Λ λ ψ, ( ) K'tn λ ψ, ( )⋅= 3( )

Equation (3) is implied in Chart 4.3 or reference 3.



Chart 4.3 from Reference 3

References

1. Walter D. Pilkey, "Formulas for Stress, Strain and Structural Matrices", Second Edition. ISBN 0-471-03221-2, Pages 285, and 330 Through 332.2. Rudolph Earl Peterson, "Stress Concentration Factors", First Edition. ISBN 9780471683292, Page 152.3. Walter D. Pilkey and Deborah F. Pilkey "Peterson's Stress Concentration Factors", Third Edition.



ORIGIN 1= ψ'1

ψ=

C

e= ψ 1.504= ψ'

1

ψ0.665=:=

Gross SCF Function

cg ψ( )

2.99690.0090

ψ−

0.01338

ψ2

+

0.12170.5180

ψ+

0.5297

ψ2

−

0.55650.7215

ψ+

0.6153

ψ2

+

4.04826.0146

ψ+

3.9815

ψ2

−

≡ cg ψ( )

2.997

0.232

1.309

6.287

=

K'tg λ ψ, ( ) cg ψ( )1

cg ψ( )2λ⋅+ cg ψ( )

3λ

2⋅+ cg ψ( )4λ

3⋅+≡

Nominal SCF Function

cn ψ( )

2.986

2.825−

2.446

=cn ψ( )

2.9890.0064

ψ−

2.872−0.095

ψ+

2.3480.196

ψ+

≡ cn ψ( )

2.985

2.809−

2.478

=

K'tn λ ψ, ( ) cn ψ( )1

cn ψ( )2λ⋅+ cn ψ( )

3λ

2⋅+≡

Determine the relationship between Kt λ ψ, ( ) and K'tn λ ψ, ( )

σmax K'tn λ ψ, ( ) σNom⋅= 4( )

σNom Λ λ ψ, ( ) σg⋅= 5( )

Substitute equation (5) into (4):

σmax K'tn λ ψ, ( ) Λ λ ψ, ( ) σg⋅( )⋅=Therefore,

σmax Kt λ ψ, ( ) σg⋅= 6( )

Where Kt λ ψ, ( ) Λ λ ψ, ( ) K'tn λ ψ, ( )⋅=


Mathcad P-elements linear versus nonlinear stress 2014-t6

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